The correct options are A x = π20 + nπ5, n ∈ I B x = π40 + nπ10, n ∈ I C x = 3π20 + nπ5, n ∈ I D x = 3π40 + nπ10, n ∈ I cos 15x=si ...

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cos 15x = sin...

Question

$cos 15 x = sin 5 x$ if

x = - \frac{π}{20} + \frac{n π}{5}, n \in I

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x = \frac{π}{40} + \frac{n π}{10}, n \in I

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x = \frac{3 π}{20} + \frac{n π}{5}, n \in I

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x = - \frac{3 π}{40} + \frac{n π}{10}, n \in I

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Solution

The correct options are
A $x = - \frac{π}{20} + \frac{n π}{5}, n \in I$
B $x = \frac{π}{40} + \frac{n π}{10}, n \in I$
C $x = \frac{3 π}{20} + \frac{n π}{5}, n \in I$
D $x = - \frac{3 π}{40} + \frac{n π}{10}, n \in I$
$cos 15 x = sin 5 x$
$\Rightarrow sin (\frac{(4 n + 1) π}{2} - 15 x) = sin 5 x$
$\Rightarrow \frac{(4 n + 1) π}{2} - 15 x = 5 x$
$\Rightarrow 20 x = 2 n π + \frac{π}{2}$
$\Rightarrow x = \frac{n π}{10} + \frac{π}{40}, n \in I$
$x = \frac{(n + 1) π}{10} - \frac{π}{10} + \frac{π}{40}$
$= \frac{k π}{10} - \frac{3 π}{40}, k = n + 1 \in I$
$cos 15 x = sin 5 x$
$\Rightarrow cos 15 x = cos (\frac{(4 n + 3) π}{2} + 5 x)$
$\Rightarrow 15 x = \frac{(4 n + 3) π}{2} + 5 x$
$\Rightarrow 10 x = 2 n π + \frac{3 π}{2}$
$x = \frac{n π}{5} + \frac{3 π}{20}, n \in I$
$x = \frac{(n + 1) π}{5} - \frac{π}{5} + \frac{3 π}{20}$
$= \frac{k π}{5} - \frac{π}{20}, k = n + 1 \in I$

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cos15x=sin5x if

$cos 15 x = sin 5 x$ if